Verified Solution • 2025–26 Session moderate

QJEEM19Ma99eb: JEE Main 2019 Mathematics - Class 12 JEEM19Ma99eb Solution: If and g(x) = log e x, (x > 0) then the value of integral is

If f(x)=2−xcos⁡x2+xcos⁡xf(x) = {{2 - x\cos x} \over {2 + x\cos x}} and g(x) = logex, (x > 0) then the value of integral

∫−π4π4g(f(x))dx\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {g\left( {f\left( x \right)} \right)dx{\rm{ }}} is

Multiple Choice Options

  • A loge3
  • B loge2
  • C loge1
  • D logee

Step-by-Step Verified Solution

g(f(x))g\left( {f\left( x \right)} \right) = ln⁡(f(x))\ln \left( {f\left( x \right)} \right) = ln⁡(2−xcos⁡x2+xcos⁡x)\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)

∴\therefore I = ∫−π4π4ln⁡(2−xcos⁡x2+xcos⁡x)dx\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right)dx}

( Using property ∫−aaf(x)dx=∫0a(f(x)+f(−x))dx\int\limits_{ - a}^a {f\left( x \right)} dx = \int\limits_0^a {\left( {f\left( x \right) + f\left( { - x} \right)} \right)} dx )

I = ∫0π4(ln⁡(2−xcos⁡x2+xcos⁡x)+ln⁡(2+xcos⁡x2−xcos⁡x))dx\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( {{{2 - x\cos x} \over {2 + x\cos x}}} \right) + \ln \left( {{{2 + x\cos x} \over {2 - x\cos x}}} \right)} \right)dx}

= ∫0π4(ln⁡(1))dx\int\limits_0^{{\pi \over 4}} {\left( {\ln \left( 1 \right)} \right)dx}

= 0 = loge1

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